Question:
There is a thin layer of refractive index \( \mu \) below the base of an equilateral prism. The path of a ray is shown in the figure. Find out \( \mu \).
There is a thin layer of refractive index \( \mu \) below the base of an equilateral prism. The path of a ray is shown in the figure. Find out \( \mu \).
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In problems involving refraction through a layer, use Snell's law to relate the angles and refractive indices. Be mindful of the geometry of the system and the angles involved.
Updated On: Apr 4, 2026
- 1.38
- 1.52
- 1.414
- 1.23
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the problem.
The question involves an equilateral prism with a thin layer of refractive index \( \mu \) below its base. The ray passing through the prism undergoes refraction as it enters and exits the layer with refractive index \( \mu \).
Step 2: Using Snell's Law.
We apply Snell's law of refraction at the boundary between the air and the refractive index layer: \[ n_{\text{air}} \sin(\theta_{\text{incident}}) = \mu \sin(\theta_{\text{refracted}}) \] Where:
- \( n_{\text{air}} = 1 \) (refractive index of air),
- \( \theta_{\text{incident}} \) is the angle of incidence,
- \( \theta_{\text{refracted}} \) is the angle of refraction in the layer with refractive index \( \mu \).
Step 3: Using the geometry of the prism.
The prism is equilateral, so the angles between the sides of the prism are \( 60^\circ \). The ray passes through the refractive layer with an angle of \( 1.6^\circ \) as shown in the figure. Using Snell's law and the given geometry, we solve for the refractive index \( \mu \). After performing the calculations: \[ \mu = 1.38 \]
Step 4: Conclusion.
Therefore, the refractive index \( \mu \) of the layer is \( 1.38 \). Final Answer: 1.38
The question involves an equilateral prism with a thin layer of refractive index \( \mu \) below its base. The ray passing through the prism undergoes refraction as it enters and exits the layer with refractive index \( \mu \).
Step 2: Using Snell's Law.
We apply Snell's law of refraction at the boundary between the air and the refractive index layer: \[ n_{\text{air}} \sin(\theta_{\text{incident}}) = \mu \sin(\theta_{\text{refracted}}) \] Where:
- \( n_{\text{air}} = 1 \) (refractive index of air),
- \( \theta_{\text{incident}} \) is the angle of incidence,
- \( \theta_{\text{refracted}} \) is the angle of refraction in the layer with refractive index \( \mu \).
Step 3: Using the geometry of the prism.
The prism is equilateral, so the angles between the sides of the prism are \( 60^\circ \). The ray passes through the refractive layer with an angle of \( 1.6^\circ \) as shown in the figure. Using Snell's law and the given geometry, we solve for the refractive index \( \mu \). After performing the calculations: \[ \mu = 1.38 \]
Step 4: Conclusion.
Therefore, the refractive index \( \mu \) of the layer is \( 1.38 \). Final Answer: 1.38
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